A ball is thrown in the air vertically with a velocity of 112 feet per second. The height above the groundseconds after release is modeled by h(t) = -16t² + 112t + 6.
a. What height was the ball originally thrown from?
b. When will the ball reach 130 feet?
c. Will the ball ever reach 250 feet? Explain.
d. When will the ball hit the ground?

a. To find the height the ball was originally thrown from, we need to look at the equation h(t) = -16t² + 112t + 6. The initial height is represented by the constant term, which is 6. Therefore, the ball was originally thrown from a height of 6 feet.

b. To find when the ball will reach 130 feet, we need to set h(t) = 130 and solve for t. This gives us the equation -16t² + 112t + 6 = 130. Simplifying, we get -16t² + 112t - 124 = 0. Dividing by -4, we get 4t² - 28t + 31 = 0. Using the quadratic formula, we find that t ≈ 1.16 seconds or t ≈ 1.84 seconds. Therefore, the ball will reach a height of 130 feet after approximately 1.16 seconds or 1.84 seconds.

c. To determine if the ball will ever reach 250 feet, we need to look at the maximum height the ball will reach. The maximum height is given by the vertex of the parabolic equation h(t) = -16t² + 112t + 6. The t-coordinate of the vertex is given by -b/2a, where a = -16 and b = 112. Therefore, t = -112/(2*-16) = 3.5 seconds. Substituting t = 3.5 seconds into the equation, we get h(3.5) = -16(3.5)² + 112(3.5) + 6 ≈ 222. Therefore, the ball will not reach a height of 250 feet.

d. To find when the ball will hit the ground, we need to set h(t) = 0 and solve for t. This gives us the equation -16t² + 112t + 6 = 0. Dividing by 2, we get -8t² + 56t + 3 = 0. Using the quadratic formula, we find that t ≈ 0.07 seconds or t ≈ 7.93 seconds. Since the ball was thrown upwards, we can discard the negative solution. Therefore, the ball will hit the ground after approximately 7.93 seconds.